A 3 kg steel ball and 2.5 m cord of negligible mass make up a simple pendulum th

Question

A 3 kg steel ball and 2.5 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point O. This pendulum is released from rest in a horizontal position and when the ball is at its lowest point it strikes a 3 kg block sitting at rest on a shelf. Assume that the collision is perfectly elastic and take the coe?cient of friction between the block and shelf to be 0.63. The acceleration of gravity is 9.81 m/s

1.What is the velocity of the block just after impact?
2.How far does the block move before coming to rest?
Answer in units of m

Explanation / Answer

      The mass of the steel ball, m = 3 kg       The length of the cord, l = 2.5 m       The coefficient of friction between the block and shelf , = 0.63 _________________________________________________________________       At its highest point, the ball will have potential energy is                    (P.E)i= mgh                               = (3 kg)(9.81 m/s2)(2.5 m)                               = 73.575 J       At its highest point, the kinetic energy of the ball is zero. Such that                    (K.E)i= 0 J _________________________________________________________________       At lowest point, the potential energy is                        (P.E)f = 0       At lowest point, the kinetic energy of the ball is                        (K.E)f = (1/2)mv2       Using the law of conservation of energy,               (P.E)i+(K.E)i =(P.E)f+(K.E)f                73.575 J+0 = 0 + (1/2)mv2                  73.575 J = (1/2)(3 kg)v2          The velocity of the ball is                  v = 7.00 m/s ____________________________________________________       Since collision is elastic, all momentum is transfered to second       object.
      Here the both masses are equal, so velocity gets interchanged                   mv+0 = mv'+0                          v' = v                              = 7.00 m/s _______________________________________________________ _______________________________________________________       The kinetic energy of the second block is                K.E' = (1/2)mv'2
                       = (1/2)(3 kg)(7.00 m/s)2                        = 73.5 J _______________________________________________________       Work done against friction till it stops is                    W = FfS       Here S is thedistance where it stops. _______________________________________________________       The frictional force, Ff = Fn       Here, the normal reaction, Fn= mg       Therefore, the frictional force equation becomes,                        Ff = mg       Therefore, the work done is given by                       W = mgS       So, the distance traveled by the block is              S = W/mg                 = (73.5 J)/(0.63)(3 kg)(9.81 m/s2)                 = 3.96 m
      Using the law of conservation of energy,               (P.E)i+(K.E)i =(P.E)f+(K.E)f                73.575 J+0 = 0 + (1/2)mv2                  73.575 J = (1/2)(3 kg)v2          The velocity of the ball is                  v = 7.00 m/s ____________________________________________________       Since collision is elastic, all momentum is transfered to second       object.
      Here the both masses are equal, so velocity gets interchanged                   mv+0 = mv'+0                          v' = v                              = 7.00 m/s _______________________________________________________ _______________________________________________________       The kinetic energy of the second block is                K.E' = (1/2)mv'2
                       = (1/2)(3 kg)(7.00 m/s)2                        = 73.5 J _______________________________________________________       Work done against friction till it stops is                    W = FfS       Here S is thedistance where it stops. _______________________________________________________       The frictional force, Ff = Fn       Here, the normal reaction, Fn= mg       Therefore, the frictional force equation becomes,                        Ff = mg       Therefore, the work done is given by                       W = mgS       So, the distance traveled by the block is              S = W/mg                 = (73.5 J)/(0.63)(3 kg)(9.81 m/s2)                 = 3.96 m
                 73.575 J = (1/2)(3 kg)v2          The velocity of the ball is                  v = 7.00 m/s ____________________________________________________       Since collision is elastic, all momentum is transfered to second       object.
      Here the both masses are equal, so velocity gets interchanged                   mv+0 = mv'+0                          v' = v                              = 7.00 m/s _______________________________________________________ _______________________________________________________       The kinetic energy of the second block is                K.E' = (1/2)mv'2
                       = (1/2)(3 kg)(7.00 m/s)2                        = 73.5 J _______________________________________________________       Work done against friction till it stops is                    W = FfS       Here S is thedistance where it stops. _______________________________________________________       The frictional force, Ff = Fn       Here, the normal reaction, Fn= mg       Therefore, the frictional force equation becomes,                        Ff = mg       Therefore, the work done is given by                       W = mgS       So, the distance traveled by the block is              S = W/mg                 = (73.5 J)/(0.63)(3 kg)(9.81 m/s2)                 = 3.96 m                 = (73.5 J)/(0.63)(3 kg)(9.81 m/s2)                 = 3.96 m

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